The New Method / Substep / Concept Idea Thread

[Cheese11]

What about:

2x2x3 on DB
EO in that orientation
solve 1 front F2L pair and build another front F2L pair on U
insert pair while solving CO and CP (~130 algs)
EP5 as on http://athefre.110mb.com/Step5.html

in STM, this would be extremely efficient (~45 STM)

Efficient yes, but I think the recognition would suck.

 

[Joël]

Hello everybody,

I had this idea for a new substep for a beginners method a while ago. I don't know if it exists already, of if it has a name. It's meant to be a small improvement to the basic beginners LBL method, a bit like working corners, except this one doesn't reduce movecount, but it makes the moves more comfortable. I succesfully taught a collegue of mine (who already knew a basic LBL method), in a few minutes. He picked it up rather well, and got results (better times) quickly. You only need 2 algs, and they are quite easy to mirror to solve pairs from to different slots.

With the standard LBL method, you'll solve the bottom layer corners oriented correctly, this results in 4 F2L pairs that are not really that nice. So I would suggest solving F2L like this:

1. Cross
2. Solve bottom layer corners, without caring about orientation, maybe even preferring the wrong orientation (=TFL? Twisty First Layer?). Ideally, this should be done with L, U and R without rotating the cube.
3. Solve 4 F2L pairs using moves like RUR'U' RUR', RU'R'U RU'R', and mirrors to solve the same cases for FL, BL and BR slots. In a case where the edge is already in the slot, do one of the two algs to insert a dummy, and use the alg that will keep the corner unoriented. (=TF2L?)


Pro's:
- Placing the corners is easier. Having a first layer corner 'upside down' (D face sticker on U face) is now the easier case.
- More comfortable 2-gen algs for the pairs, resulting in less rotations during 2nd layer.
- A small step from standard LBL towards solving pairs.

Con's:
- 2 more algs to learn

 

[Kirjava]

^
That Roux variation sound pretty similar to a variation of Waterman that Waterman himself proposed. You solve the first layer minus an edge followed by CLL. I've tried it out before. I always try to insert an edge for the second Roux block whole solving the top corners.


New (?) Roux Grip
I've always used the grip as described by Kirjava in his thread, but I used my pinky for single M moves and I overlay my fingers on the E slice over the D layer.

A while ago, a had a couple friends who I saw using doing H and Z perms with a different grip. They were alternating using there ring fingers to preform the M moves.

The grip they used is good for the permutations, but it doesn't give you freedom with the U layer for improvising while doing Roux L6E. From there grip I came up with a similar one. Since I'm more used to using my pinky than my ring finger, I held the two blocks with my ring fingers and thumbs while preforming the M moves with my two pinkies alternating. This frees up your pointer and middle finger on the left and right hand for doing double-flicks. This helps incase it is necessary to do two U2 turns (separated by an M move of course) because you can alternate which hand you double-flick with.

Uhm. How do you do M?

 

[elrog]

Uhm. How do you do M?

With a 57mm cube I use either one of my pinkies to push the bottom side of the BD edge up. On my 50mm cube, I will do a single M move the same way, but if I have 2 Ms in a row, (separated by a U move of course) I will do the first M move the way I have already described and for any consecutive M moves afterwards I will bring my other pinkie forward and push the FD edge back.

 

[thesolver]

Hey guys,
Just thought of a concept for 5x5 and above.
Here it is.

Solve the first 8 edges using freeslices.
Put the 8 edges on the top and bottom.
Then use some algs to solve the remaining 4 edges. Probably would take around 10 to 30 algs.
Please tell me if someone has already developed this idea before or if any of you already use this.

 

[BoBoGuy]

Hey guys,
Just thought of a concept for 5x5 and above.
Here it is.

Solve the first 8 edges using freeslices.
Put the 8 edges on the top and bottom.
Then use some algs to solve the remaining 4 edges. Probably would take around 10 to 30 algs.
Please tell me if someone has already developed this idea before or if any of you already use this.

Otherwise known as...*bum bum bum* REDUCTION!

 

[Suratha]

A (New?) Petrus method variant:Rotten 2x2x2 block
Step-1:Solve the 2x2x2 block;the corner can be twisted-so fewer moves than normal 2x2x2.
Step-2:Finish 2x2x3.
Step-3:Get all the wrong edges to the top layer(0-3 moves).If six edges are wrong,get four on top.We can solve the twisted corner and edge orientation using a single alg and finish the F2L.If there are no wrong edges,finish the F2L and do CLS or CPLS(fewer cases because the corner is in place).
Step-4LL or 2-gll.
Questions:
i)Is this exists anywhere else? If yes,please forgive me.If no,give some opinions about this variant.
ii)Will it be faster than the normal Petrus method?

 

[Ninja Storm]

Otherwise known as...*bum bum bum* REDUCTION!

He's talking about solving all four edges at once. With algs.


@thesolver: I don't know how good this would be. I feel like the last four edges would always have to be the same, which would limit the effectiveness of the first eight edges. Could be possible, though.

 

[TheOneOnTheLeft]

Hey guys,
Just thought of a concept for 5x5 and above.
Here it is.

Solve the first 8 edges using freeslices.
Put the 8 edges on the top and bottom.
Then use some algs to solve the remaining 4 edges. Probably would take around 10 to 30 algs.
Please tell me if someone has already developed this idea before or if any of you already use this.

You'd almost certainly have to do y2/x2 for recognition, which would be slow. Also there are 10 cases for last 2 edges, so I think there's be a lot more than 30 algs for last 4 edges. I could be wrong on that though.

 

[cubingboss]

New Method

I thought of this new method, but I am not sure how well it will work.
Step 1: Orient all edges
Step 2: Make a 1x2x3 block
Step 3: Make a 1x2x3 block in the middle layer
For this step it does not matter what side it is made on but you must place it on the bottom
Step 4: Make a 1x2x3 blcok on the side that is left
Step 5: Do last layer however you want if you have the F2L matched up
If not Solve the corners. Then finish the left and right side.
I will try to upload a walkthrough of this so I can show better how this works.
If you have any suggestions please tell me.

 

[Ollie]

so...a less efficient version of Petrus?

 

[TDM]

I thought of this new method, but I am not sure how well it will work.
Step 1: Orient all edges
Step 2: Make a 1x2x3 block
Step 3: Make a 1x2x3 block in the middle layer
For this step it does not matter what side it is made on but you must place it on the bottom
Step 4: Make a 1x2x3 blcok on the side that is left
Step 5: Do last layer however you want if you have the F2L matched up
If not Solve the corners. Then finish the left and right side.
I will try to upload a walkthrough of this so I can show better how this works.
If you have any suggestions please tell me.

If you swap steps 2 and 3, that's ZZ with NMLL.

 

[cubingboss]

It's sort of petrus and roux ish.

 

[Phillip1847]

This idea is (in my mind) very simple, and rather obvious perhaps:
1. EO
2. F2l however you like - last slot
3. Insert last slot while simultaneously permuting the already oriented edges.
4. L4C(thanks tnfeliks)

Just for fun, I created some algs for step 3 + example solve. Recognition is a pain.

https://docs.google.com/document/d/1iT8xk068YPDTgfRRoG_jGugdVxGC3eIDyZ_O8LUFXes

 

[Bunyanderman]

This idea is (in my mind) very simple, and rather obvious perhaps:
1. EO
2. F2l however you like - last slot
3. Insert last slot while simultaneously permuting the already oriented edges.
4. COLL, which solves the cube.

This idea is so simple that I'm sure someone else already made it.

Step 4 would need to be zbll, the edges would get un-permuted if you used regular cases.
Step 3 would be inefficient, might as well put the last pair in, and permute after. Even then oll/pll would do the job for you.

 

[TheNextFeliks]

This idea is (in my mind) very simple, and rather obvious perhaps:
1. EO
2. F2l however you like - last slot
3. Insert last slot while simultaneously permuting the already oriented edges.
4. COLL, which solves the cube.

This idea is so simple that I'm sure someone else already made it.

So basically phasing+ZZLL for cases where edges are permuted. L4C is the real name for step 4. It's on the wiki.

 

[Phillip1847]

Yes, except you always phase the entire thing.
I guess its inefficient. The shortest "alg" is 4 moves long and the longest one is 9.
Just for fun, I actually found the algorithms and recog. Its in the original post.

Thanks for the correction.

 

[elrog]

To me, this seems kind of life a simplified Heise last slot. This is also what I did intuitively while building up to Heise. Heise is more efficient because you solve another corner while doing it ending with L3C, but its not good for speedsolving.

I have merged my next post with this one to fix a bug.

[FONT=&amp]I was trying to think of an easier system for recognizing CLL. I don't really like the standard CLL recognition which is why I have not learned any form of CLL yet. I thought about learning CLL using hyperorientation which also allows for non-matching CLL, but then I realized you had to recognize the orientation case along with where the right and left colors are, meaning that if you want to recognize for non-matching CLL, you're going to have 2 corners that have a top color different than the other two and it really messes things up.

I then thought to recognize CLL using only 2 adjacent colors. You could pick these colors by always looking for 2 particular adjacent colors, or just by looking at a single corner (such as the top left) and seeing where the other stickers are that match its 2 colors (not counting the top color). So essentially, you recognize the CLL case with 4 stickers. You will need to distinguish which color is which when finding the two adjacent colors (the corner that had both colors you’re looking for doesn’t count). I would recommend looking for the right/left side sticker first on the corner that shares both. If you ever move on to non-matching CLL, you will want to know if your corner is a left or right sided corner (unless of course, you always use the same corner and it is always on the opposite side of the cube from what you do your setup move on; for example, you always use the blue/red/white corner which is on your left side and you only have your block/corners off by R moves). If your doing non-matching CLL, you must remember to substitute in your new adjacent color.

This sounds great, but the biggest drawback (in my opinion) is the fact that you may have to look at another side (such as the back side) to see a sticker you need to see (keep in mind that you should actually just AUF, not do a rotation). This is also true of standard CLL recognition though, and you can already know without having to look at another side if you saw the corner you were looking for and tracked it (though you can only do this if you always pick the same 2 adjacent colors).

One other thing some people may see as a problem is the fact that you have to learn multiple recognition cases for each algorithm. I am completely fine with this because it just means you also are recognizing the angle you are looking at the case from so you’ll know how to AUF. If someone added up all the different angles to recognize PLL from 2-sides, you’d have a lot more than 21. It will also be easy to remember the recognition cases because you only need to remember the orientation of a single corner + 2 more stickers. There are, however, 162 different possibilities for the placement of the stickers you’re looking for.

I calculated the 162 like this: If you always use the top left corner, it has 3 orientation. One of its matching stickers will be on one of the other 3 corners at any of 3 different orientations. The last sticker you’re looking for will be on either of the 2 remaining corners at any of 3 different orientations. This gives you 3 x (3x3) x (2x3) = 162. Now this may seem odd because if you take the 42 CLL cases and multiply by 4 for recognizing from any of the 4 corners, you get 42 x 4 or 168. This means that you either have 6 cases that are unrecognizable without more information, some CLL cases look the same from multiple corners, or a combination of the two. I’m fairly sure that this is just some cases looking the same with different corners (the H orientation case with corners permuted is an example of this and accounts for 2 of the 6).

You still may have different top colors for non-matching CLL recognition and you may have to find the color that is the equivalent of a colors adjacent color rather than that color itself. I also think that you would want to always pick the same two adjacent colors for non-matching CLL.[/FONT][FONT=&amp][/FONT][FONT=&amp]
[/FONT][FONT=&amp]
I’d like to know if you think I’ve made any mistakes and any thoughts you have about this method of CLL recognition. Name suggestions perhaps?[/FONT]

 

[Michael Womack]

I was trying to think of an easier system for recognizing CLL. I don't really like the standard CLL recognition which is why I have not learned any form of CLL yet. I thought about learning CLL using hyperorientation which also allows for non-matching CLL, but then I realized you had to recognize the orientation case along with where the right and left colors are, meaning that if you want to recognize for non-matching CLL, you're going to have 2 corners that have a top color different than the other two and it really messes things up.

I then thought to recognize CLL using only 2 adjacent colors. You could pick these colors by always looking for 2 particular adjacent colors, or just by looking at a single corner (such as the top left) and seeing where the other stickers are that match its 2 colors (not counting the top color). So essentially, you recognize the CLL case with 4 stickers. You will need to distinguish which color is which when finding the two adjacent colors (the corner that had both colors you’re looking for doesn’t count).

This sounds great, but the biggest drawback (in my opinion) is the fact that you may have to look at another side (such as the back side) to see a sticker you need to see (keep in mind that you should actually just AUF, not do a rotation). This is also true of standard CLL recognition though, and you can already know without having to look at another side if you saw the corner you were looking for and tracked it (though you can only do this if youalways pick the same 2 adjacent colors).

One other thing some people may see as a problem is the fact that you have to learn multiple recognition cases for each algorithm. I am completely fine with this because it just means you also are recognizing the angle you are looking at the case from so you’ll know how to AUF. If someone added up all the different angles to recognize PLL from 2-sides, you’d have a lot more than 21. It will also be easy to remember the recognition cases because you only need to remember the orientation of a single corner + 2 more stickers. There are, however, 162 different possibilities for the placement of the stickers you’re looking for.

I calculated the 162 like this: If you always use the top left corner, it has 3 orientation. One of its matching stickers will be on one of the other 3 corners at any of 3 different orientations. The last sticker you’re looking for will be on either of the 2 remaining corners at any of 3 different orientations. This gives you 3 x (3x3) x (2x3) = 162. Now this may seem odd because if you take the 42 CLL cases and multiply by 4 for recognizing from any of the 4 corners, you get 42 x 4 or 168. This means that you either have 6 cases that are unrecognizable without more information, some CLL cases look the same from multiple corners, or a combination of the two. I’m fairly sure that this is just some cases looking the same with different corners (the H orientation case with corners permuted is an example of this and accounts for 2 of the 6).

You still may have different top colors for non-matching CLL recognition and you may have to find the color that is the equivalent of a colors adjacent color rather than that color itself, but this is still better than hyperorientation (in my opinion). I also think that you would want to always pick the same two adjacent colors for non-matching CLL.

I’d like to know if you think I’ve made any mistakes and any thoughts you have about this method of CLL recognition. Name suggestions perhaps?

sounds alot like this TCLL http://www.speedsolving.com/forum/showthread.php?44438-TCLL-New-2x2-Method

 

[elrog]

TCLL is a kind of CLL that orients some of the bottom corners at the same time as orienting and permuting the top corners. What I came up with is not a kind of CLL. It is a recognition system for CLL.